PLEASE HELP THIS IS MY LAST QUESTION! Triangle JKL is an equilateral triangle with two of its vertices at points J and K. What are the coordinates of point L? Round to the nearest tenth as needed.

(6, 7.2)

Step-by-step explanation:

Since it is an equilateral triangle, all of the side length must be equal.

Therefore, we know the lengths JK = 6 and JL = 6. If h is the height of the triangle, the Pythagoras's theorem says

(JK/2)² + h² = JL²

putting in the values of JK and JL gives

(6/2)² + h² = 6²

3² + h² = 6²

subtracting 3^2 from both sides gives

h² = 6² - 3²

h² = 27

Taking the square root of both sides gives

h =

h = 5.2

With the value of h in hand, we can now read off the coordinates of L.

The x coordinates of L is 6 (count the boxes along the x-axis until you are under L or halfway between J and K).

The y-coordinates of L is 2 + h = 2 + 5.2 = 7.2 ( how far above the x-axis the triangle is plus the height of the triangle).

answer: 1 + 1 = 11

step-by-step explanation:

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15